[YOOHYOJEONG] WEEK 1 Solutions

contains-duplicate/YOOHYOJEONG.py

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+# https://leetcode.com/problems/contains-duplicate/
+
+class Solution(object):
+    def containsDuplicate(self, nums):
+        if len(nums) > len(set(nums)):
+            return True
+        else:
+            return False

house-robber/YOOHYOJEONG.py

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+# https://leetcode.com/problems/house-robber/
+
+class Solution(object):
+    def rob(self, nums):
+
+        prev = 0
+        prev_2 = 0
+
+        for num in nums:
+            cur = max(prev, prev_2 + num)
+
+            prev_2 = prev
+            prev = cur
+
+        return prev

longest-consecutive-sequence/YOOHYOJEONG.py

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+# https://leetcode.com/problems/longest-consecutive-sequence/
+
+class Solution(object):
+    def longestConsecutive(self, nums):
+
+        nums_set = set(nums)
+        lens = 0
+
+        for n in nums_set:
+
+            if n-1 not in nums_set:
+                cur = n
+                lenth = 1
+
+                while cur+1 in nums_set:
+                    cur += 1
+                    lenth += 1
+
+                lens = max(lens, lenth)
+
+        return lens

top-k-frequent-elements/YOOHYOJEONG.py

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+# https://leetcode.com/problems/top-k-frequent-elements/
+
+class Solution(object):
+    def topKFrequent(self, nums, k):
+        cnt = {}
+        for num in nums:
+            if num in cnt:
+                cnt[num] += 1
+            else:
+                cnt[num] = 0
+
+        sorted_cnt = sorted(cnt.items(), key=lambda x: x[1], reverse=True)
+
+        return [item[0] for item in sorted_cnt[:k]]

two-sum/YOOHYOJEONG.py

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+# https://leetcode.com/problems/two-sum/description/
+
+# class Solution(object):
+#     def twoSum(self, nums, target):
+#         for i in range(len(nums)):
+#             for j in range(len(nums)):
+#                 if nums[i]+nums[j] == target:
+#                     if i != j:
+#                         return [i, j]                    
+# > 해당 방식 사용 시 시간 복잡도가 O(n²)이라 개선을 해 보고자 아래 솔루션으로 재풀이 진행
+
+class Solution(object):
+    def twoSum(self, nums, target):
+        seen = {}
+
+        for i, num in enumerate(nums):
+            diff = target - num
+
+            if diff in seen:
+                return [seen[diff], i]
+
+            seen[num] = i
+
+# index를 기억하도록 하면 반복문 한번 O(n), 딕셔너리 조회 평균 O(1)로 전체 O(n)이 됨